# Homework Help: Determine magnitude of force acting on block on inclined plane

1. Sep 26, 2010

1. The problem statement, all variables and given/known data
A 20-kg package is at rest on an incline when a force P is applied to it. Determine the magnitude of P if 10s is required for the package to travel 5m up the incline. The static and kinetic coefficients of friction between the package and the incline are both equal to 0.3.

Answer: 419N to start and 301 during sliding

2. Relevant equations
m=20kg
g=9.8m/s2
t=10s
xi = 0ft
xf = 5ft
μs = μk = 0.3
Frictional force = Ff = μN
3. The attempt at a solution
First, I drew a free-body diagram including tilted coordinate axes:

Then, I went about finding the sum of the forces equations in the x- and y-directions.

$$\sum F_{x}=-F_{f}-mgsin(20)+Pcos(30)=ma$$

$$\sum F_{y}=N-mgcos(20)-Psin(30)=0$$

From the second equation,
$$N=mgcos(20)+Psin(30)$$

Plugging this value for N into Ff = μN in the first equation yields:

$$P=\frac{m\left(a+\mu gcos(20)+gsin(20)\right)}{cos(30)-\mu sin(30)}$$

If I know the acceleration, I can find the magnitude of P. There has to be a way to find acceleration from the initial conditions, but I'm at a loss. Also, once static friction is overcome, I'm sure the sum of the forces equations will be different but I'm not sure how.

2. Sep 26, 2010

### Thefox14

Well since the force seems to be constant, you could apply the kinematic equation to solve for the acceleration:
$$x = x_0 + v_0t + (1/2)at^2$$

3. Sep 27, 2010

From the $\sum F_{x}$ equation, along with $F_{f}=\mu N$:

$$a=-\mu gcos(20)-\frac{\mu Psin(30)}{m}-gsin(20)+Pcos(30)$$

$$x_{f}=x_{i}+v_{i}t+\frac{1}{2}at^{2}$$

$$5=\frac{1}{2}\left (-\mu gcos(20)-\frac{\mu Psin(30)}{m}-gsin(20)+Pcos(30) \right )\left (10^{2} \right )$$

$$\frac{1}{10}=-\mu gcos(20)-\frac{\mu Psin(30)}{m}-gsin(20)+Pcos(30)$$

$$\frac{1}{10}+\mu gcos(20)+gsin(20)=P\left (cos(30)-\frac{\mu sin(30)}{m} \right )$$

$$P=\frac{\frac{1}{10}+\mu gcos(20)+gsin(20)}{cos(30)-\frac{\mu sin(30)}{m}}=\frac{\frac{1}{10}+(0.3)(9.8)cos(20)+(9.8)sin(20)}{cos(30)-\frac{0.3sin(30)}{20}}\approx 7.24N \neq \: Book's\: answer\: of\: 419N$$

Did I do the math wrong? Where do I go from here?